Examples from First-order methods

Exponential growth with an initial value

For

the initial value is nonzero, so separate on the branch $y\neq0$:

Therefore

Since $y(0)=1>0$, the positive branch is selected and the constant is zero:

A lost equilibrium and a real-domain restriction

For

the nonconstant solutions on a real branch satisfy

Integration gives

The square root requires $|y|\leq1$. Because $\arcsin y$ takes values in $[-\pi/2,\pi/2]$, the implicit relation describes a branch only while $x^2/2+C$ stays in that range.

The constant solutions $y=1$ and $y=-1$ were lost on division. At a point where a nonconstant branch reaches either level, it can also join a constant branch: the right side is zero there and uniqueness is not guaranteed at $|y|=1$.

A family of circles

For

multiply by $y\,dx$ and integrate:

After absorbing the factor of $2$ into the constant,

The solution curves are circles, but the original differential equation is undefined when $y=0$. Thus a circle is represented by an upper or lower graph branch on an interval that does not cross its points on the $x$-axis. The initial points $(0,7)$, $(2,3)$ and $(-6,-8)$ select respectively

An equation already in product-rule form

For

the left side is already $(e^{x^2}y)'$. Hence

and

A standard first-order linear equation

For

the integrating factor is $\mu=e^{3x}$. Multiplication gives

Therefore

so

A singular normal form with a smooth continuation

For

divide by $x$ only on $x>0$ or $x<0$:

On either interval, an integrating factor is

Multiplying by a nonzero constant does not change an integrating factor, so we may take $\mu=xe^x$ on each chosen interval. Then

Thus

The normalised equation is valid only for $x\ne0$, but the original equation at $x=0$ still imposes $y(0)=2$. Choosing $C=-2$ gives

by continuous extension. Other values of $C$ do not extend continuously through $x=0$, so the interval restriction created by division must still be reported.

A polynomial potential

For

we have

so the equation is exact. Integrate $M$ with respect to $x$:

Then

which gives $h'(y)=2y-7$ and hence $h(y)=y^2-7y$. The implicit general solution is

A potential involving hyperbolic functions

Rewrite

as

Here

Integrating $M$ with respect to $x$ gives

Matching

gives $h(y)=y^4/4$. Therefore

A degree-zero right side

For

solve for the derivative:

The right side is homogeneous of degree $0$. Work on an interval where $x\ne0$ and $y\ne0$, so the displayed normal form and the substitution $v=y/x$ are valid. Put $y=xv$, so $y'=v+xv'$. Substitution gives

and, after division by $x^2$,

Separate and integrate:

Exponentiating gives $(1+3v^2)|x|^3=C_1$. The sign of $x$ is fixed on the chosen interval, so that sign can be absorbed into the constant. Substituting $v=y/x$ gives the equivalent implicit solution

on a chosen interval with $x\neq0$ and $y\neq0$. Differentiating this relation gives

which is equivalent to the original equation.

A Bernoulli equation with a zero solution

For

use $v=y^{-4}$ on a nonzero branch. Since $v'=-4y^{-5}y'$, multiplying by $y^{-5}$ and rearranging gives

The integrating factor is $x^2$, so

Thus

or, on intervals where the right side gives a real nonzero solution,

These real branches require $x\neq0$ and $-4x^3+C/x^2>0$. The solution $y=0$ is also present in the original equation but is not represented by $y^{-4}$.